Question: You have found the following ages (in years) of all 6 porcupines at your local zoo: $ 3,\enspace 1,\enspace 24,\enspace 9,\enspace 23,\enspace 6$ What is the average age of the porcupines at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{3 + 1 + 24 + 9 + 23 + 6}{{6}} = {11\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-8$ years $64$ years $^2$ $1$ year $-10$ years $100$ years $^2$ $24$ years $13$ years $169$ years $^2$ $9$ years $-2$ years $4$ years $^2$ $23$ years $12$ years $144$ years $^2$ $6$ years $-5$ years $25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{64} + {100} + {169} + {4} + {144} + {25}} {{6}} $ $ {\sigma^2} = \dfrac{{506}}{{6}} = {84.33\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{84.33\text{ years}^2}} = {9.2\text{ years}} $ The average porcupine at the zoo is 11 years old. There is a standard deviation of 9.2 years.